如何在URL中的最后一个斜杠之后获得所有内容？(How to get everything after last slash in a URL?)

如何在URL中的最后一个斜杠之后获得所有内容？(How to get everything after last slash in a URL?)

如何在Python中的URL中提取最后一个斜杠的内容？ 例如，这些URL应该返回以下内容：

URL: http://www.test.com/TEST1 returns: TEST1 URL: http://www.test.com/page/TEST2 returns: TEST2 URL: http://www.test.com/page/page/12345 returns: 12345

我尝试过urlparse，但是这给了我完整的路径文件名，比如page/page/12345 。

How can I extract whatever follows the last slash in a URL in Python? For example, these URLs should return the following:

URL: http://www.test.com/TEST1 returns: TEST1 URL: http://www.test.com/page/TEST2 returns: TEST2 URL: http://www.test.com/page/page/12345 returns: 12345

I've tried urlparse, but that gives me the full path filename, such as page/page/12345.

最满意答案

您不需要花哨的东西，只需看到标准库中的字符串方法 ，您就可以轻松地在“文件名”部分和其余部分之间分隔您的URL：

url.rsplit('/', 1)

所以你可以通过以下方式获得感兴趣的部分：

url.rsplit('/', 1)[-1]

You don't need fancy things, just see the string methods in the standard library and you can easily split your url between 'filename' part and the rest:

url.rsplit('/', 1)

So you can get the part you're interested in simply with:

url.rsplit('/', 1)[-1]